Question

A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displacement? (the shear modulus for aluminum is 2.5 X 1010N/m2).

a) 4.4 X 10-8 m
b) 8.2 X 10-8 m
c) 1.9 X 10-8 m
d) 5.3 X 10-8 m

Answer 1

The resultant relative displacement caused by a shear force applied to an aluminum cube is calculated using the shear modulus of aluminum. The resulting displacement, given the values provided, is 5.3 x 10¹¸ m, which corresponds to option (d).

Step-by-step explanation:

The question is about calculating the resultant relative displacement caused by a shear force applied to an aluminum cube using the shear modulus. Using the formula for shear strain (γ = Δx/h), where δx is the displacement, h is the height, and knowing that the shear stress (τ = F/A), which equals to shear modulus (S) times shear strain (γ), we can find the relative displacement. To find the displacement, rearrange the formula to Δx = (τ/S) * h.

Given the values:

• Shear force (F) = 400 N
• Side of the cube (h) = 30 cm = 0.3 m (for both height and side, since it's a cube)
• Shear modulus (S) for aluminum = 2.5 x 10¹¹ N/m²

The area (A) onto which the force is applied is one face of the cube: A = (0.3 m)² = 0.09 m².

Using the shear stress formula: τ = F/A = 400 N / 0.09 m² = 4444.44 N/m²

Now find the displacement: Δx = (τ/S) * h = (4444.44 N/m²) / (2.5 x 10¹¹ N/m²) * 0.3 m = 5.3 x 10¹¸ m

Therefore, the relative displacement of the aluminum cube is 5.3 x 10¹¸ m, corresponding to option (d).

Answer 2

`\Delta x=5.33* 10^(-8)m`

So option (d) will be correct option

Step-by-step explanation:

We have given shear force F = 400 N

Size of the cube
`L_0=30cm=0.3m`

Shear modulus of aluminium
`S=2.5* 10^(10)N/m^2`

We have to find the resulting relative displacement

Area of the cube
`A=L_0^2=0.3^2=0.09m^2`

We know that shear force is given by

`F=S* (\Delta x)/(L_0)* A`

So
`400=2.5* 10^(10)* (\Delta x)/(0.3)* 0.09`

`\Delta x=5.33* 10^(-8)m`

So option (d) will be correct option